Prime number goes into 1s

dgreenb801

Show that for any prime $p>5$ , it goes into the number consisting only of $p-1$ 1s.
For example,
$7|111111$
$11|1111111111$
$13|111111111111$
$17|1111111111111111$
etc.

**Posted:** Sun Nov 01, 2009 11:49 am

Kouichi Nakagawa

$9|10^{p - 1}$ is trivial.
$p|10^{p - 1}$ is true. Because Fermat's little theorem.
Thus $9p|10^{p - 1} \iff p|\underbrace{11\cdots11}_{p - 1}$ .

**Posted:** Sun Nov 01, 2009 1:13 pm

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JBL

You probably want to think those claims over some more Wink

**Posted:** Sun Nov 01, 2009 1:45 pm

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Joel
Hi Deeps! <3

veezbo

BLAH. It took me way too long to solve this problem.

Solution

**Posted:** Sun Nov 01, 2009 4:18 pm

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diophantient

**Posted:** Tue Nov 03, 2009 4:55 pm

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JBL

You don't have a problem with the claim that 9 divides a power of 10? Rolling Eyes

**Posted:** Tue Nov 03, 2009 5:05 pm

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Joel
Hi Deeps! <3

dgreenb801

Of course, he meant $9p|10^{p - 1} - 1$ .

**Posted:** Tue Nov 03, 2009 6:40 pm

Kouichi Nakagawa

Sorry.
I missed words slightly.
As dgreenb801 says definitely; as follows.

$9|10^{p - 1}\color{red}-1\color{black}$ is trivial.
$p|10^{p - 1}\color{red}-1\color{black}$ is true. Because Fermat's little theorem.
Thus $9p|10^{p - 1}\color{red}-1\color{black} \iff p|\underbrace{11\cdots11}_{p - 1}$ .

**Posted:** Tue Nov 03, 2009 9:20 pm

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